<p><span style="color:#27ae60;"><span style="font-size:20px;">Solving The Problem</span></span></p>
<p>This seems like a <strong><span style="color:#8e44ad;">Words to Algebra</span></strong> problem. So, we can start reading <strong><span style="color:#8e44ad;">Piece by Piece</span></strong> and convert it into equations and expressions.</p>
<blockquote>
<p>A rectangular garden </p>
</blockquote>
<p>We know the garden will have a length and a width, so let's set up some variables:</p>
<ul>
<li>$L$ = The length of the garden</li>
<li>$W$ = The width of the garden</li>
</ul>
<blockquote>
<p>has a perimeter of 92 meters.</p>
</blockquote>
<p>The perimeter is twice the sum of the length and the width, so:</p>
<p>$2(L+W)=92$</p>
<blockquote>
<p>If the length of the garden is 1 meter greater than twice its width,</p>
</blockquote>
<ul>
<li>$L=2W+1$</li>
<li>$W=\frac{L-1}{2}$</li>
</ul>
<blockquote>
<p>what is the length of the garden, in meters?</p>
</blockquote>
<p>So, we have to solve for the length $L$. We can substitute our new equation for $W$ (in terms of $L$) into the first to figure out $L$:</p>
<ul>
<li>$2(L+\frac{L-1}{2}) = 92$</li>
<li>$2L+2\frac{L-1}{2} = 92$</li>
<li>$2L+L-1= 92$</li>
<li>$3L-1=92$</li>
<li>$3L=93$</li>
<li>$L=\frac{93}{3}$</li>
<li>$L=31$</li>
</ul>
<p>So, the answer is <font color="#27ae60">D</font>.</p>